# How do you find the zeros of y=x^3-10x-12?

Jan 26, 2018

Zeros are $x = - 2 , x = - 1.65 \mathmr{and} x = 3.65$

#### Explanation:

$y = {x}^{3} - 10 x - 12$ or

$y = {x}^{3} + 2 {x}^{2} - 2 {x}^{2} - 4 x - 6 x - 12$ or

$y = {x}^{2} \left(x + 2\right) - 2 x \left(x + 2\right) - 6 \left(x + 2\right)$ or

$y = \left(x + 2\right) \left({x}^{2} - 2 x - 6\right) \therefore x = - 2$ is one zero

${x}^{2} - 2 x - 6 = 0 \mathmr{and} {x}^{2} - 2 x = 6$ or

${x}^{2} - 2 x + 1 = 6 + 1$ or

${\left(x - 1\right)}^{2} = 7 \mathmr{and} \left(x - 1\right) = \pm \sqrt{7}$ or

$x = 1 \pm 2.65 \therefore x \approx 3.65 , x = - 1.65$

$\therefore y = {x}^{3} - 10 x - 12 = \left(x + 2\right) \left(x + 1.65\right) \left(x - 3.65\right)$

Zeros are $x = - 2 , x = - 1.65 \mathmr{and} x = 3.65$ [Ans]